Discriminant of Quintic (Galois theory)?

Question

I am working on the irreducible polynomial $x^5-Npx+p=0$ where $p$ is a prime and $N\in\mathbb Z_+$. I need to calculate the discriminant of this to determine its Galois group, background here by Conrad, here by Yuan who explains facts about Galois groups of irreducible quintics and a specific discriminant question here. The general formula for Discriminant by Wikipedia is

$$\triangle=a_n^{2n-2}\prod_{i<j}(r_i-r_j)^2$$

where $a_n$ is the leading coefficient and $r_i,\ldots,r_n$ are the roots. In the example, $a_5=1,a_1=-Np,a_0=p$ and otherwise $a_i=0$. For roots

$$x(x^4-Np)=-p$$

where I don't understand yet how to solve $x$ so unable to get the discrimiant. So

How to calculate the discriminant of an irreducible quintic?

Answer 1

Just for reference: the discriminant of your polynomial is $$3125 p^4-256 N^5 p^5.$$ I computed it using Mathematica. For comparison, the discriminant of the general monic polynomial $x^5+a x^4+b x^3+c x^2+d x+e$ is $$256 a^5 e^3-192 a^4 b d e^2-128 a^4 c^2 e^2+144 a^4 c d^2 e-27 a^4 d^4+144 a^3 b^2 c e^2-6 a^3 b^2 d^2 e-80 a^3 b c^2 d e+18 a^3 b c d^3-1600 a^3 b e^3+16 a^3 c^4 e-4 a^3 c^3 d^2+160 a^3 c d e^2-36 a^3 d^3 e-27 a^2 b^4 e^2+18 a^2 b^3 c d e-4 a^2 b^3 d^3-4 a^2 b^2 c^3 e+a^2 b^2 c^2 d^2+1020 a^2 b^2 d e^2+560 a^2 b c^2 e^2-746 a^2 b c d^2 e+144 a^2 b d^4+24 a^2 c^3 d e-6 a^2 c^2 d^3+2000 a^2 c e^3-50 a^2 d^2 e^2-630 a b^3 c e^2+24 a b^3 d^2 e+356 a b^2 c^2 d e-80 a b^2 c d^3+2250 a b^2 e^3-72 a b c^4 e+18 a b c^3 d^2-2050 a b c d e^2+160 a b d^3 e-900 a c^3 e^2+1020 a c^2 d^2 e-192 a c d^4-2500 a d e^3+108 b^5 e^2-72 b^4 c d e+16 b^4 d^3+16 b^3 c^3 e-4 b^3 c^2 d^2-900 b^3 d e^2+825 b^2 c^2 e^2+560 b^2 c d^2 e-128 b^2 d^4-630 b c^3 d e+144 b c^2 d^3-3750 b c e^3+2000 b d^2 e^2+108 c^5 e-27 c^4 d^2+2250 c^2 d e^2-1600 c d^3 e+256 d^5+3125 e^4,$$

Answer 2

There is no need to calculate the discriminant to find this Galois group if $N \geq 2$. Let $f(x) = x^5 - Npx + p$ with $N \geq 2$ and let $G$ be the Galois group of $f(x)$. We have \begin{equation*} \begin{aligned} f(-Np) &= -N^5p^5 + N^2p^2 + p < 0\\ f(0) &= p > 0 \\ f(1) &= 1 + p - Np < 0. \end{aligned} \end{equation*} Since $f(x)$ has two sign changes, it must have at least two real roots. Since the derivative $f'(x) = 5x^4 - Np$ has only two real roots, it follows that $f(x)$ cannot have all real roots. But the nonreal roots come in conjugate pairs, and so $f(x)$ must have three real roots and two nonreal roots. Therefore the transposition which interchanges the two nonreal roots is in $G$, hence $G$ is isomorphic to $S_5$.

If $N = 1$ then it depends on how large $p$ is. You can show that if $p \geq 13$ then there are three real roots, and then again $G$ is isomorphic to $S_5$. Now if $p < 13$, you can check by graphing that there is only one real root and then $G$ contains a $5$-cycle and a double transposition and is thus isomorphic to either $A_5$ or $D_5$.

Returning to the question about the discriminant, let $f(x) = x^5 + ax + b$ be an irreducible quintic. Let $\alpha_1,\ldots,\alpha_5$ be the complex roots of $f(x)$. Then $f(x) = (x - \alpha_1)\cdots(x - \alpha_5)$, and so taking the derivative with the product rule gives $$f'(x) = \sum_{i=1}^{5}\prod_{\substack{1 \leq j \leq 5\\ i\neq j}}(x - \alpha_j),$$ hence $$f'(\alpha_i) = \prod_{\substack{1 \leq i,j \leq 5\\ i\neq j}}(\alpha_i - \alpha_j)$$ You can check that $\Delta = \prod_{i=1}^{5}f'(\alpha_i)$. We also have $f'(x) = 5x^4 + a$, hence $f'(\alpha_i) = 5\alpha_i^4 + a$. Therefore $$\Delta = \prod_{i=1}^{5}(5\alpha_i^4 + a).$$ Expanding the above product out gives a formula for $\Delta$ in terms of the elementary symmetric polynomials in $\alpha_1^4,\ldots,\alpha_5^4$. Reexpressing these in terms of the elementary symmetric polynomials in $\alpha_1,\ldots,\alpha_5$ thus gives a formula for $\Delta$ in terms of the coefficients of $f(x)$.